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I am read a solution (4.9) Here say:

... both $a, d$ are pure phases, so that it is always possible to find (non unique) real numbers $\alpha, \beta, \delta$ such that $a = e^{i(\alpha−\beta/2−\delta/2)}$...

My question: What is a pure phases? Why $a, d$ are pure phase?. Why it is always possible to find (non unique) real numbers $\alpha, \beta, \delta$ such that $a = e^{i(\alpha−\beta/2−\delta/2)}$?

...Therefore it is possible to assume that none of the cœfficients $a, b, c, d$ vanish. In particular, there are real numbers $0 < γ, γ^{'} < \pi$ such that $|a| = \cos γ/2$

Why the module of any Complex number can be represented than $\sin$ or $\cos$ functions?

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  • $\begingroup$ I would understand a "pure phase" to be a complex number of absolute value 1. So it has the form $e^{i\phi}$ for some angle $\phi$. $\endgroup$ – Andreas Blass Apr 16 '13 at 0:22
  • $\begingroup$ all complex number can be represent in the form $e^\phi$? $\endgroup$ – juaninf Apr 16 '13 at 0:25
  • $\begingroup$ I mean real angles. $\endgroup$ – Andreas Blass Apr 16 '13 at 0:52
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    $\begingroup$ @juaninf All complex numbers can be represented as $ae^{ib}$ for real $a$ and $b$: this is the polar form of a complex number. The ones with $a=1$ form the unit circle in the complex plane. $\endgroup$ – rschwieb Apr 16 '13 at 1:05

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